STRUCTURE OF B-9 AND B-10
By Prof.Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which cannot lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Boron (B''') naturally occurs in two isotopes, 10B and 11B, the latter of which makes up about 80% of natural boron. In the following diagrams of the structure of B-9 and B-10 one sees that the symmetry and the total spins are the two fundamental characteristics which lead to the nuclear structure of them. '''NUCLEAR STRUCTURE OF THE B-9 WITH S = -3/2 In the diagram of B9 you see that p1, n1 and p2 have a total spin S = -3/2 which gives the total spin S = -3/2 of the nucleus. It consists of the smaller rectangle having the nucleons p1,n1,p3,n3, p4, and n4 ,which form a rectangle similar to the rectangle of the isolated Li6 with a binding energy B(Li) = -31.98 MeV. (See my STRUCTURE AND BINDING OF Li6 AND Li5 ). Since the binding energy of the B9 is B(B9) = - 56.31 MeV one should conclude that the nucleons outside the structure of Li6 would give extra pn bonds with an extra binding energy Bex . Such an energy is given by (Bex + Uex) = B(B9) – B(Li6) = (-56.31) – (-31.98) = -24.33 MeV This equation could tell us that the rectangle (p1n1n4p4) is stronger than the B(Li6) = -31.98 MeV of the isolated Li6. However here the number of protons is greater than that of the lithium and at the corners the bonds per proton are only two per proton. For example the p1 makes only the two bonds like p1n1 and p1n3 which cannot overcome the pp repulsions of long range. ' DIAGRAMS OF THE UNSTABLE B-9 AND STADLE B-10' Boron-9 with S = -3/2 Boron-10 with S = +3 p4 (-1/2).n4 (- 1/2).p5(-1/2) n5(+1/2)p3(+1/2).n3(+1/2)p5(+1/2) n3(+1/2).p3(+1/2).n2(+1/2) n2(-1/2)..p2( -1/2) p1(-1/2). n1(-1/2).p2(-1/2) n4(+1/2)p1(+(1/2).n1(+1/2)p4(+1/2) NUCLEAR STRUCTURE OF B-10 WITH S = +3 In the case of the stable structure of B-10 after a careful analysis of the diagram of B-10 one sees that the rectangle (p1n1n3p3 ) represents the structure of the isolated Li6. So outside the rectangle of Li6 we see the nuclei p4 and n4 with a total spin S = +1/2 +1/2 = +1 and having the bonds p1n4 and n1p4 in radial direction. That is, each of them seems to have the structure of deuteron. In the same way we see the p5((+1/2) and the n5(+1/2) which make horizontal bonds with n3 and p3 respecitively. This situation leads to the stability of B-10 because the bonds at the corners are not two per proton ( as in the case of the unstable B9) but three per proton. For example in the stable B-10 the p1 makes the three bonds like p1n1, p1n4, and p1n2. Also the magnetic attraction of the p4 and p5 of parallel spin along the z axis contributes to the increase of the binding energies of the pn bonds. Here we see 8 nucleons of positive spins and 2 nucleons of negative spins Thus the total spin of B-10 is S = +8/2 -2/2 = +6/2 = +3. IMPORTANT CONCLUSIONS: It is of interest to note that the rectangle having the structure of Li6 contributes significantly to the binding energy B(B-10) = - 64.75 MeV, while the radial pn bonds of nucleons p4, n4 , p5 and n5 have the structure of deuterons with weak energies. It means that the binding energy of the rectangle is stronger than the B(Li6) = - 31.38 MeV of the isolated Li6. It is not surprising because at points p1 and n1 there exist three pn bonds per nucleon, which contribute to the increase of the binding energy, while in the isolated Li6 there exist always two pn bonds per nucleon. For example at p1 of B-10 there exist the bonds p1n1, p1n2, p1n4 . Category:Fundamental physics concepts